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Woler Ono Consliuct 95% confidence interval (or Ine populalion proportion olc Hhham M10.0D in retirees in Ine United Slales 25 ycars older who have less savings:Int...

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Woler Ono Consliuct 95% confidence interval (or Ine populalion proportion olc Hhham M10.0D in retirees in Ine United Slales 25 ycars older who have less savings:Interpret Ine inlerval:louncpe 29 and themean simple random sample of size n is drawn. The sampre sample stcndad deviation; founc ope Consinuct 90% confidence interval for u | Ine sample sizeConslmuct = 90% confidence interva itc | the sample size;I0O.How dces Increosing Ine sample size aftect the margin emor; E?Doynolma curve and label

woler Ono Consliuct 95% confidence interval (or Ine populalion proportion olc Hhham M10.0D in retirees in Ine United Slales 25 ycars older who have less savings: Interpret Ine inlerval: lounc pe 29 and the mean simple random sample of size n is drawn. The sampre sample stcndad deviation; founc ope Consinuct 90% confidence interval for u | Ine sample size Conslmuct = 90% confidence interva itc | the sample size; I0O. How dces Increosing Ine sample size aftect the margin emor; E? Doy nolma curve and label the mean and inileciion points: 25 and 0 =7



Answers

Construct the appropriate confidence interval. A simple random sample of size $n=40$ is drawn from a population. The sample mean is found to be $\bar{x}=120.5$ and the sample standard deviation is found to be $s=12.9 .$ Construct a $99 \%$ confidence interval about the population mean.

Following a solution to number 12, and this gives a summary stats that the sample main X bar is 35.1 and the sample standard deviation was a 8.7. Now notice it's a sample the esses sample standard deviation, and we don't know what sigma is, the population standard deviation. So since we don't know sigma, we have to use the tea interval instead of the z interval. Remember the less, you know, the more variable is going to be. So the t interval is what we're gonna use here, since we know s not sigma, and we're asked to find three things and then we talk about Um a little bit of conditions for inference on this last part, but let's look at a B&C because they kind of go together first, we're gonna find the 90% confidence interval winter and is 40. Then we're going to find the same level of confidence, 90% confidence interval, but this time the end is 100. So we increase that sample size And then this 3rd part We increase the confidence of them, but we keep the N equal to 40 again. And then we're going to talk a little bit about the conditions for inference. Now in the beginning it did not say anything about The population distribution. So normally it says assume that the population distribution is approximately normal. Well, here we don't need it to be normal because N. Is large enough. That magic number is 30. As long as your sample size is at least 30. It doesn't matter how your population is distributed. Um Because that's just the role as the central limit theorem world. Alright, so I'm gonna use technology for this because it cuts down. But you can use any excel. I'm gonna use the T. I. T. For but you can use any sort of technology Wish. And if you'd like to use the formula you should get the same answers. But I'm gonna use T. I. D. Four. So if you go to stat and then air over the tests it's this eighth option here the T. Interval. So you can scroll down or you can just click eight and make sure the summary stats is highlighted. So we don't have any data. Um we don't have a long list of data or anything. We're just giving the summary stats. So keep that highlighted and it asks for X. Bar. That's 35.1. S. Was 8.7 in was 40 at least for this first one and then the sea levels 90% of 900.9 For this first one. And we're gonna go and calculating this first band here. That's our interval. So 32.8 To 37.4 is r. t. interval. So 37 point I'm sorry 30-32 point 782 To 37.418. Okay so that's kind of like our base. So it's about uh we can kind of estimate maybe about five a part of 4 4.7 units apart. So just kind of keep that in the back of your mind 4.7 units apart. So let's see what happens whenever we increase that sample size to 100 and you may be able to guess by now. So if we go to the T. Interval again this time I'm just going to change this into 100 And let's see what happens. So we go from 36 I'm sorry 33.6-36.5. So let's write that down. Let's compare 33 .655. All the way up to 36 545 Okay. So notice that the lower bound gets higher and the upper bound gets lower. That means remember appear we were up to 4.7 here. It's You know about three actually a little bit less than 3-9. That interval um is Only 2.9 units so it's gotten narrower. So the next part of this question is how does the sample size or how does how do the degrees of freedom influence that margin of error? Or intern how does it influence the interval with? So as in increases or as the degrees of freedom increase, that margin of error is going to decrease. Which makes so making the confidence interval more narrow. Okay, so as you increase your in and you can kind of think about that standard air that? S over square devin, you're dividing by a bigger number which makes that margin of error smaller um and the smaller the margin of air then more narrow your confidence interval will be. So now let's look at the 98% confidence interval. But this time the ends back to 40. So if we go to stat Tests and it's that 8th 1 And this time we're going to bring that down to 40 and then we're gonna bring this one up to 98% of .98. We calculate and we get 31- 38. 31.8- 38.4. So it's right that down 31 point 763 two, points 437. So we're gonna compare this one to part a where it has the same sample size, just different confidence level. So we increase the confidence level and notice that we went from 4.7 units apart. And this is like 6.7 units apart. So it's gotten wider. All right. So What's that relationship? Well, as the confidence level increases, which is what we did here went from 90 to 98. That margin of air also increases because that intervals wider. So making the confidence interval wider. So the more confident you are, the wider and wider that interval must be. And then part D. It says what conditions of inference are needed if N equals 18. So if N equals 18 and I kind of sort already hinted on this, But since that end is less than 30, the population must be normally distributed since And is less than 30. Okay, so since N is less than 30, That population needs to be normal. Now. Normally they say that in the beginning but this time they didn't because these sample sizes are big enough, it doesn't matter. But if that in were equal to 18, 1 of those conditions for inference, if it is less than 30 is that the population comes from a normal distribution.

The following is a solution to # seven. And this says that a random sample of 12 people were selected. We don't really know what it is, but it's from a normal approximation or a normal distribution approximately normal. So that allows us to have this small sample size. Normally, you have to have a sample size of at least 30. But since it comes from a normal distribution, we can just say that this is okay, so in equals 12 is fine. And the uh the sample mean X bar was 45 And the Sample Standard Deviations 14. And we're asked to find the 90% confidence interval for the population mean. So, since we're estimating the population mean, we're either going to use the Z or the tea interval, and we're gonna use the tea interval, because we don't know what sigma is, we don't know that population standard deviation, because we weren't given given that we were only given the sample standard deviation. S so because of that, we have to use the T interval. Okay, so terrible. So the 90% confidence interval now you can use the tea interval formula, you can certainly go about it that way. I'm gonna use the form of the calculator just because it goes by a little bit more quickly. But if you go to stat and then tests, it's gonna be this eighth option down here, this is on a. T. I. T. For uh T. Interval, so I'm gonna click eight and make sure summary stats is Is highlighted there and I'm just gonna start filling in my stuff. So x bar is 45. The sample standard deviation is 14 And the n. was 12 And we want to be 90%. So I want to say .9 there for the sea level And we calculate and then this top band here gives us what we need. So 37 742 and 52.258. Let's go and write that down. So 37 742 All the way up to 52 258 It doesn't say to do this, but if it asks for an interpretation, you would just say we are 90% confident that the true population mean mu is between 37.742 and 52.258.

Problem. 22. The critical values using table six four degrees of freedom equal to end minus one, which is 30 minus one, is equal 29. So the chi square off one minus Alfa over two is equal to 14.257 and the chi square or all five people to 49.588 So the boundaries for the standard deviation is n minus one over Chi Square of Alfa over to DR as is equal to 6.256 and the other boundary in minus one over Chi square, off one minus Alfa over to dot ESC is equal to 11.667 The other boundaries for the variants are the square value off these values, which is 39.14 and 136.12

The following is a solution for # three. And we're asked to find a 980% confidence interval given this this data here. So we're gonna use the formula that I have up in red here, it's p one minus p two hats. In this case it's 20.255 -0.193 plus or minus Now the Z score here. If you don't have it memorized, that's fine. In fact, I can show you how to get it real quick because this one isn't as typical. So 1995-99 are usually the most typical ones. But if we go to distribution here and inverse norm And we type in .8 And then the center will be center is always for confidence intervals and then we paste, we get positive negative 1.282. So that's what we're gonna use for our z value. So positive, negative 1.282 times the square root of 0255 Times 1 -255. Which is 0.745 Divided by that sample size of 300 Plus 019, three Times 1- that. Which is 0807. And that's divided by the other sample size of 400 and you should get 0.062 plus or -0.041. So you can leave it like that. Or you can actually expand it out Where you subtract and you add you should get 0.021. Oops, Okay. All the way up to 0.103. Okay, so either answer is acceptable. Now we find a 95% confidence interval Given this following data. So we started out the same way now 95% confidence that we should know that Z value that's 196. But let's go and start this .1 47 which is the P one hat minus 10.131 Plus or -1.96 which is the critical value for 95 times the square root of .147 Times 1 -147 which is 853. And then divide that by the sample size of 3500 And then plus .131 Times 1 -131, which is 869. And divide that by the sample size of 37 50. So whenever you do that you should get .016 plus or minus .0159. So you can leave it like this certainly. Or you can expand it out groups Where you get essentially zero, also point 0001 All the way up to 0.0319. Okay, so those are two representations of the 95% confidence interval. Given this data


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